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Q & A: H emission spectrum

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Most recent answer: 10/22/2007
Can you tell me the wavelength of the EM wave that would correspond to the photon emitted when a hydrogen atomís electron falls from the n2 to the n1 energy level?

Thank you very much,

- adam waltzer
seattle, WA USA
Sure, thatís a standard problem. The energy of light is given by E=pc, where p is the momentum and c is the speed of light. For anything, p=h/lambda, where lambda is the wavelength an h is Planckís constant.  So E= hc/lambda for photons, or lambda= hc/E. hc happens to be 1240 eV-nm.
So now all we have to do is figure out the energy given off in that n=2 to n=1 transition. The energies of those states turn out to be -13.6 eV/n2. Therefore the n=2 level is 10.2eV higher than the n=1 level. So lambda is 1240/10.2 nm or 122nm.
This particular wavelength is called the Lyman alpha transition and is used by astronomers and cosmologists to determine properties of the early universe. 

Mike W.

Lee H

(published on 10/22/2007)

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